A single tree can be placed in the following way: root the tree in an arbitrary node, and place that node on an arbitrary node of the hypercube (say ~0~). Then we do a DFS on the tree, and when moving from a tree node that is placed on the hypercube node ~x~ to a new node using the ~i^\text{th}~ edge, we place it on the hypercube node ~x \oplus 2^i~.

The rest of the trees can be placed as follows: for each ~x \in \{0, \dots, 2^n-1\}~ that has an even number of ones in binary, we take the hypercube nodes on which the first tree was placed and xor their labels with ~x~. Notice that in such a way, we obtain ~2^{n-1}~ trees.

A proof that the described trees make a tiling is left to the reader.